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Có 2 cách giải:
\(xy+2x+3y+5=0\)
\(\Leftrightarrow x\left(y+2\right)=-3y-5\)
\(\Leftrightarrow x=\frac{-3y-5}{y+2}\)
\(\Leftrightarrow x=\frac{-3y-6}{y+2}+\frac{1}{y+2}\)
\(\Leftrightarrow x=-3+\frac{1}{y+2}\)
Để \(x\in Z\)
Mà \(-3\in Z\)
\(\Rightarrow\frac{1}{y+2}\in Z\)
\(\Rightarrow1⋮\left(y+2\right)\)
\(\Rightarrow\orbr{\begin{cases}y+2=-1\\y+2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=-3\\y=-1\end{cases}}\)
*Nếu y = -3 => x = - 4.
*Nếu y = -1 => x = -2.
Đề gì mà dài dữ vậy !? Nhìn đã thấy choáng rồi =_=
Đề 3 bài 5 :
Ta đặt vế trái là A
Vì \(xyz=2006\)
=>A= \(\dfrac{xyzx}{xy+xyzx+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z=1}\)
=> \(\dfrac{zx}{1+zx+x}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}=1\)
=> đpcm
Đề 4 bài 5 :
Ta có : \(a+b=c+d\)
\(\Leftrightarrow\left(a+b\right)^2=\left(c+d\right)^2\)
\(\Leftrightarrow a^2+2ab+b^2=c^2+2cd+d^2\)
\(\Leftrightarrow2ab=2cd\) ( Vì \(a^2+b^2=c^2+d^2\))
\(\Leftrightarrow a^2-2ab+b^2=c^2-2cd+d^2\)
\(\Leftrightarrow\left(a-b\right)^2=\left(c-d\right)^2\)
Xét hai trường hợp :
TH1: \(a-b=c-d\)
Mà ta có : \(a+b=c+d\)
\(\Rightarrow a-b+a+b=c-d+c+d\)
\(\Leftrightarrow2a=2c\)
\(\Leftrightarrow a=c\) \(\Rightarrow b=d\) (*)
TH2: \(a-b=d-c\)
Mà \(a+b=c+d\)
\(\Leftrightarrow a-b+a+b=d-c+d+c\)
\(\Leftrightarrow2a=2d\)
\(\Leftrightarrow a=d\) \(\Rightarrow b=c\) (**)
Thay vào....
Từ (*)và (**) => đpcm
P/s : Làm hộ mấy bài thôi ,dài quá mỏi tay :vv
Bài 2:
1)a) x2 + 3x + 3y + xy
= x(x + 3) + y(x + 3)
= (x + 3)(x + y)
b) x3 + 5x2 + 6x
= x(x2 + 5x + 6)
= x(x2 + 2x + 3x + 6)
= x[x(x + 2) + 3(x + 2)]
= x(x + 2)(x + 3)
2) Biến đổi vế trái ta có:
(x + y + z)2 - x2 - y2 - z2
= x2 + y2 + z2 + 2(xy + yz + xz) - x2 - y2 - z2
= 2(xy + yz + xz)
= vế phải
⇒ đccm
2. Rút gọn biểu thức: \(\left(x+y\right)^2-\left(x-y\right)^2-4\left(x-1\right)y\)
Giải:
\(\left(x+y\right)^2-\left(x-y\right)^2-4\left(x-1\right)y\)
=> \(x^2+y^2+2xy-x^2-y^2+2xy-4xy+4y\)
=> 4y
Bài 1: (1,5 điểm)1. Làm phép chia: (x2+ 2x + 1) : (x + 1)
=> (x2+ 2x + 1) : (x + 1)
=> \(\left(x+1\right)^2:\left(x+1\right)\)
=> x+1
Bài 1:
a) 2x(x2 - 3x + 4)
= 2x3 - 6x2 + 8x
b) (x + 2)(x - 1)
= x2 - x + 2x - 2
= x2 + x - 2
c) (4x4 - 2x3 + 6x2) : 2x
= 2x3 - x2 + 3x
Bài 2:
a) 2x2 - 6x
= 2x(x - 3)
b) 2x2 - 18
= 2(x2 - 9)
= 2(x - 3)(x + 3)
c) x3 + 3x2 + x + 3
= x2(x + 3) + (x + 3)
= (x + 3)(x2 + 1)
Bài 1 :
a) \(2x\left(x^2-3x+4\right)\)
= \(2x^3-6x^2+8x\)
b) \(\left(x+2\right)\left(x-1\right)\)
\(=x^2-x+2x-2\)
\(=x^2-x-2\)
Bài 2 :
a) \(2x^2-6x\)
\(=2x\left(x-3\right)\)
b) \(2x^2-18\)
\(=2\left(x^2-9\right)\)
\(=2\left(x-3\right)\left(x+3\right)\)
c) \(x^3+3x^2+x+3\)
\(=\left(x^3+3x^2\right)\left(x+3\right)\)
\(=x^2\left(x+3\right)\left(x+3\right)\)
\(=\left(x^2+1\right)\left(x+3\right)\)
Bài 3 :
a) \(\dfrac{5x}{x-1}+\dfrac{-5}{x-1}=\dfrac{5x+\left(-5\right)}{x-1}=\dfrac{5\left(x-1\right)}{x-1}=5\)
b) \(\dfrac{1}{x-3}+\dfrac{2}{x+3}+\dfrac{9-x}{x^2-9}\)
\(=\dfrac{1}{x-3}+\dfrac{2}{x+3}+\dfrac{9-x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{9-x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x-6+9-x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)
Bài 1 :
a) \(x^2-2x+2y-xy\)
\(=\left(x^2-2x\right)+\left(2y-xy\right)\)
\(=x\left(x-2\right)+y\left(2-x\right)\)
\(=x\left(x-2\right)-y\left(x-2\right)\)
\(=\left(x-y\right)\left(x-2\right)\)
b) \(x^2+4xy-16+4y^2\)
\(=\left(x^2-16\right)+\left(4xy+4y^2\right)\)
\(=\left(x-4\right)\left(x+4\right)+4y\left(x+y\right)\)
\(=\left(x-4\right)\left(x+4+4y\right)\left(x+y\right)\)
Bài 3 :
a) \(K=\left(\dfrac{a}{a-1}-\dfrac{1}{a^2-a}\right):\left(\dfrac{1}{a+1}+\dfrac{2}{a^2-1}\right)\)
\(K=\left(\dfrac{a^2}{a\left(a-1\right)}-\dfrac{1}{a\left(a-1\right)}\right):\left(\dfrac{a-1}{\left(a+1\right)\left(a-1\right)}+\dfrac{2}{\left(a+1\right)\left(a-1\right)}\right)\)
\(K=\left(\dfrac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}\right):\left(\dfrac{a-1+2}{\left(a+1\right)\left(a-1\right)}\right)\)
\(K=\dfrac{a+1}{a}:\dfrac{1}{a+1}=\dfrac{a+1}{a}.a+1=\dfrac{\left(a+1\right)^2}{a}\)
Để biểu thức K được xác định thì \(a\ne0\)
b) Với \(a=\dfrac{1}{2}\) thay vào biểu thức ta có :
\(K=\dfrac{\left(\dfrac{1}{2}+1\right)^2}{\dfrac{1}{2}}=4,5\)
và bằng
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